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good point

I have 10 distinguishable socks in my drawer: 4 white, 4 brown, and 2 blue. In how many ways can I choose a pair of socks, provided that I get two brown socks or two white socks?

There are 4*3 = 12 ways to get two brown socks and

there are 4*3 = 12 ways to get two white socks

so that is 24 ways.

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melody and guest.

I wasn't targeting a cheat nerdiest.

I have no more reason to think that this person is cheating than I do for most others.

I simply like to encourage people to contribute to their own solutions.

They will be more litkely to learn that way.

If they do not want to do that, well fair enough, but if they should not be help until they interact.

it's real hard to read your question.

can you properly write latex?

just copy the text inside your dollar signs and go to the latex button. then, paste the text you copied into the text box for latex and then click the OK button and voila!

Is this what you mean? and if so, what did you do to start this problem? where is your work?

\(n = 2^{10} \cdot 3^{18} \cdot 5^{4}\)

nice job targeting cheats. :) you can recognize them right away with their - i was proven WROng by the great melody.

helllllllllllllllllllllllllp 

ASAP

Due tomorrow

and stuff like that.

Ah. Thank you.

What have you done towards answering this for yourself?

Please show us.

\(Let \qquad ki=\frac{z_2}{z_1} \qquad k\in Z\\ \displaystyle \left| \frac{2z_1+7z_2}{2z_1-7z_2}\right|\\ =\displaystyle \left| \frac{\frac{2z_1+7z_2}{z_1}}{\frac{2z_1-7z_2}{z_1}}\right|\\ =\displaystyle \left| \frac{2+7ki}{2-7ki}\right|\\ =\displaystyle \left| \frac{(2+7ki)^2}{4+49k^2}\right|\\ =\displaystyle \left| \frac{(4-49k^2)+28ki}{4+49k^2}\right|\\ =\displaystyle \frac{\sqrt{(4-49k^2)^2+(28k)^2}}{4+49k^2}\\ =\displaystyle \frac{\sqrt{ 16+2401k^4-392k^2 +784k^2}}{4+49k^2}\\ =\displaystyle \frac{\sqrt{ 2401k^4+392k^2+16 }}{4+49k^2}\\ =\displaystyle \frac{ 49k^2+4 }{4+49k^2}\\ =1 \)

You need to check that.

Let \qquad ki=\frac{z_2}{z_1} \qquad k\in Z\\
\displaystyle \left|  \frac{2z_1+7z_2}{2z_1-7z_2}\right|\\
=\displaystyle \left|  \frac{\frac{2z_1+7z_2}{z_1}}{\frac{2z_1-7z_2}{z_1}}\right|\\
=\displaystyle \left|  \frac{2+7ki}{2-7ki}\right|\\
=\displaystyle \left|  \frac{(2+7ki)^2}{4+49k^2}\right|\\
=\displaystyle \left|  \frac{(4-49k^2)+28ki}{4+49k^2}\right|\\
=\displaystyle   \frac{\sqrt{(4-49k^2)^2+(28k)^2}}{4+49k^2}\\
=\displaystyle   \frac{\sqrt{   16+2401k^4-392k^2      +784k^2}}{4+49k^2}\\
=\displaystyle   \frac{\sqrt{   2401k^4+392k^2+16 }}{4+49k^2}\\
=\displaystyle   \frac{  49k^2+4 }{4+49k^2}\\
=1

What is the value of x?

Hello Guest!

\(b(b^4 \cdot b^5)^2 = b^x\\ b\cdot b^8\cdot b^{10} = b^x\\ b^{1+8+10}=b^x\\ b^{19}=b^x\\ \color{blue}x=19\)

  !

\(n^2-3n+2 = p\\ n^2-2n-n+2 = p\\ n(n-2)-1(n-2) = p\\ (n-2)(n-1) = p\\ \)

p is a prime equal to the product of two consecutive numbers. Since the product of two consecutive numbers is always even, p is an even prime number so p = 2.So n = 3.

Mostly we don't bother, but yes, moderators are able to label a post as 'best answer'

To make this problem easier, let's just shift the parabola up one unit so the we have the parabola  y =  x^2

The height  of our triangle = x^2  and the base =  2x

To have an area of 8  we need to  solve this

(1/2)* 2x  * x^2   = 8

x^3  =  8

x = 2

At x = 2 the base will be  2(2)  = 4  and the height will be  2^2 =  4  = y

And similarly   for and area of  64  we need  to  solve this

(1/2) * 2x  * x^2 =  64

x^4 =  64

x =  4

At x = 4, the base =  2(4) = 8   and the height = 4^2  = 16 = y

So....we  need to  shift everything down  by one  unit

y = "r"  will range from   (4-1) =  3  to  (16 -1)  =   15

So  r =   [ 3 , 15 ] 

Let the quantity in the large bag =  Q oz

Then in the bonus bag the quantity  = 1.2Q oz

Cost of large bag = $12

Cost of bonus bag = $8

Cost per oz in bonus bag  =    8  / (1.2)Q  

Cost per oz in large bag    =    12 / Q

So

[8 / (1.2 Q) ] / [ 12/ Q ] = 

8 / (1.2 Q)  *    Q / 12   =

8 / (12 * 1.2) = 

8 / 14.4  =

80 / 144  =

5 /  9

Substitute the point into the second equation: \(-4k + 7 = -7\). 

Now, just solve for k. (note that equation m isn't needed)

Note that \(-a^y = a^y\) when y is even. 

This means that all the terms with an even-powered term will cancel out. 

Thus, the resulting expansion will have \(\color{brown}\boxed{6}\) terms: \(a^{11}, a^9, a^7, a^5, a^3, a\)

I think you're right Guest, I made a mistake...

Here's another way: 

Let the width of the rectangle be \(w\), and let the length be \(l\)

The area of the original rectangle is \(w \times l\).

The area of the new rectangle is \(0.75w \times 1.25l = 0.9375 \times w \times l\).

This means that the area was decreased by \(1 - 0.9375 = \color{brown}\boxed{6.25 \text{%}}\)

                                    C

                                   33

                            E                  D

                                    O

    78     A                    F                        B 39

Let O be the center of the incircle

ECOD  forms a  quadrilateral  with angles  OEC and OED both being right

So  angle  EOD  =  180 - angle C  = 180- 33  = 147° =  minor arc ED

And angle EFD is an inscribed angle inrercepting this arc

So its measure = 147 /  2   =   73.5°

Geno answered this one!

:https://web2.0calc.com/questions/help-as-soon-as-you-can

The correct answer is 3/5.

(2 + x) / 2 =  -17                        (4 + y)  /  2 =   8

2 + x =  -17 * 2                          4 + y  = 2 * 8

2 + x = -34                                 4 + y =  16                           

x  =  -34 -2                                   y =  16 - 4

x =  -36                                         y =  12

x^2 - 4x + y^2 - 6y - 36 = 25       

x^2 - 4x + y^2 -6y  = 25 + 36       complete the square on x, y

x^2 - 4x + 4 + y^2 - 6y + 9  = 25 + 36 + 4 + 9

25 + 36 + 4 + 9 =  r^2

74   = r^2

sqrt (74 )   = r

In 2010, the sum of the year's digits is 3. What is the first year after 2010 in which the sum of the digits is 3, and the year is odd? 

The year 20,001 will work, unless you consider it to be so far in the future that it's out of contention.